Power supply + DC2DC coverter --> 4,65 Volt instead of 5 Volt

Hello,

I use this power supply:

Power: 72Watt
Output Voltage: 12V DC
Output Current: 6Amp
Supply Voltage:
120...373V DC
88...264V AC
Protection:
overload
short-circuit
voltage growth
Number of Outputs: 1

Power Supply

and at the output I have connected this dc2dc step down power converter:
Features:

  1. In+: Input positive
  2. Out+: Output positive
  3. GND: IN-/OUT-
  4. Dimension:453116mm(lengthwidthheight)
  5. Input voltage: 9-35V(Maximum limit voltage is 35V, please leave some leeway to use ,otherwise be damaged)
  6. Output voltage: 5V
  7. Output current: 5A
  8. Output power: 25W

dc2dc converter

I measured 11.1 Volt in the output of the power supply (which is also the input to the DC2DC converter)
and I measured 4,65 Volt in the output of the output of the dc2dc converter. It should show 5 Volt. I use 1,5 mm^2 cables all inside the circuit and the connections, 3 pads for the conductors in the proto board and the supply network is 220Volt/50Hz. What could be wrong??

Have you ever measured other power supplies? Your numbers are within 5% so I would guess they are in spec (although, those parts do not provide specs).

Do not expect to spend $3 on ebay and get lab-grade instruments.

Can I fix the output voltage to be 5 Volt instead 4,65 Volts, which now as it is I can't use it. Any circuit to rise the voltage, and lose in current maybe?

I believe this does work (??), because it has adjustable output:

Module properties: non isolated buck module (BUCK)

XLSEMI XL4005E1
Input voltage: 5-32V
Continuous adjustable output voltage: 1-32V
Output current: 0-5A, recommended for use in 4.5A.
Output power: it is recommended to use in the 75W, more than 50W Please add heat sink.
Working temperature: -40~+85 degrees
Operating frequency: 180KHz
96%: the highest conversion efficiency (efficiency is related to the input, the output voltage, current, pressure difference)
Load adjustment rate: S (I) is less than or equal to 0.8%
Voltage regulation rate: S (U) is less than or equal to 0.8%

The output voltage is controlled by the value of 2 resistors.

Have a look at figure 15 on page 20 of the TPS40057 datasheet.

The resistors R1 and RBIAS set the output voltage.
They form a potential divider and the circuit regulates when the voltage on pin7 becomes 0.7V.
(Note that this is a 3.3V buck converter and so the values on your board are going to be different).

You should be able to trace the track from pin7 of IC U1 to identify which 2 resistors are involved.

(They won't be called R1 and RBIAS on your board, but I can't make them out from the photograph on the seller's website).

s-l500.jpg

By changing the value of one or other of those 2 resistors you can adjust the output voltage. However you may not be able to get the voltage to be exactly 5V using standard resistor values.

Two items with a lower than expected voltage is suspicious. Test your DMM.
I have (had) a voltmeter that showed a lower voltage when the internal battery was almost flat.
Leo..

I changed dc2dc step down converter, I configured it to 5,1 Volt (and I measured with the multimeter to check it) , but when I measure the current I get 0 Amperes...and indeed the final device (the load) does not work..

Why are you worrying about a slightly lower Voltage? It isn't going to be an issue 99% of the time.

My device needs 5 Volt to work, I thought 4,65 Volt was the issue for not working, but I realized that I have bigger problems, such as: no current!

alex5678:
My device needs 5 Volt to work,

No it doesn't, or it is at least highly unlikely.

alex5678:
My device needs 5 Volt to work, I thought 4,65 Volt was the issue for not working, but I realized that I have bigger problems, such as: no current!

What do you mean "no current"
Can you tell us what you project is and what it comprises of?
Can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png?

Thanks.. Tom.... :slight_smile:

Ok, I fixed it, faulty elements...!! I bought another one. The maximum consumption is 5A * 12 Volt = 60 Watt, when stressed. Should I use a fan?

Hi,
What is the application?
What does your project do?
What is the load for the dc-dc converter?

Thanks.. Tom.... :slight_smile:

Ok. It is a circuit with sensors, CPU, IO and stuff like that. The load when stressed is 5A @ 5 Volts and the supply is 6A @ 12 Volt. Please also, see post #1. The project does heavy load processing. Thanx..